MA 724 Homework 5 – Solutions
Warm-up (do not turn in):
A. If P contains 0 on its boundary, then P ∆ will be an unbounded polyhedron. Indeed, P
has some valid inequality ax ≤ 0 with a 6= 0, so any positive multiple of a will lie in P ∆ .
If 0 ∈
/ P , then not only will P ∆ be an unbounded polyhedron, but also P ∆∆ 6= P .
(Indeed, 0 ∈ P ∆∆ but 0 ∈
/ P .)
B. Every face of P × Q is a face of P times a face of Q. Hence ∆2 × ∆2 has 9 vertices
(vi , wj ) for i = 1, 2, 3 and j = 1, 2, 3, and (vi , wj ) and (vi0 , wj 0 ) are adjacent if i = i0 or
j = j 0 . Thus its graph (indeed, its Schlegel diagram) looks like:
Any two facets of ∆2 × ∆2 are adjacent, so the graph of its polar is the complete graph
on 6 vertices (every pair of vertices is connected).
Homework:
1. (a) Clearly if cx ≤ 0 for all x ∈ C, then cx ≤ 1 for all x ∈ C. Conversely, suppose
cx ≤ 1 for all x ∈ C. Since C is closed under taking positive scalar multiples, we
must have c(tx) = tcx ≤ 1 for all t > 0. Hence cx ≤ 1t for all t > 0, so cx ≤ 0.
(b) We claim that C ∆ has H-representation given by P (Y T , 0) = {c | cY ≤ 0} and
V-representation given by cone(AT ) = cone{ai | ai ∈ A}.
For the first claim, clearly if cP∈ C ∆ , then
P cyi ≤ 0 for all yi ∈ Y . Conversely, if
cyi ≤ 0 for all yi ∈ Y , then c( λi yi ) = λi (cyi ) ≤ 0 for all λi ≥ 0, so c ∈ C ∆ .
P
P
For the second claim, if c =
λi ai ∈ cone(AT ), then cx =
λi ai x ≤ 0 for all
x ∈ C. Conversely, suppose cx ≤ 0 for all x ∈ C = P (A, 0). Then by Farkas’
Lemma, cx ≤ 0 can be written as a linear combination of the inequalities ai x ≤ 0
defining C; hence c ∈ cone(AT ).
Now C is pointed ⇔ its lineality space ker(A) is trivial ⇔ A has rank d ⇔ AT has
rank d ⇔ the row span of AT is Rd ⇔ C ∆ = cone(AT ) has dimension d.
Similarly, C = cone(Y ) has dimension d ⇔ Y and Y T have rank d ⇔ C ∆ = P (Y T , 0)
is pointed.
(c) If C is the homogenization of a polytope P , then C = cone{(1, x) | x ∈ P }. Then
(−1, c) ∈ C ∆ if and only if −1 + cx ≤ 0, that is, cx ≤ 1. Hence C ∆ intersected with
the plane x0 = −1 is P ∆ (so C ∆ is a reflection of the homogenization of P ∆ ).
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2. (a) Suppose P is reflexive and v ∈ int(P ) is a nonzero lattice point. Then some point
tv for t > 1 lies on a facet whose affine span is ax = 1, where a is a lattice point in
(Rd )∗ . But then av = 1t is not an integer, which is a contradiction.
(b) Let P ⊆ R3 be given by the inequalities xi ≤ 1, −xi ≤ 1, and x1 + x2 + x3 ≤ 2.
Then P contains no lattice points in its interior other than the origin (since the only
way for an integer to satisfy xi < 1 and −xi < 1 is if xi = 0). But P ∆ has the vertex
( 12 , 12 , 21 ), so P is not reflexive.
3. We induct on d. The case d = 1 is trivial.
Let c = (0, 0, . . . , 0, 1) ∈ (Rd )∗ . Let the minimum and maximum value of cx for x ∈ P
be m and M . Recall that if v ∈ P is a vertex for which cv is not minimal, there exists
an adjacent vertex x such that cx < cv.
Now pick any two vertices v and w, and suppose first that cv + cw ≤ m + M . Then
there exists a sequence v = v0 , v1 , . . . , va of adjacent vertices such that cvi > cvi+1 and
cva = m. Since all the values of cvi are integers, a ≤ cv − m. Similarly, we can find a
sequence w = w0 , w1 , . . . , wb of adjacent vertices such that cwb = m and b ≤ cw − m.
Since va and wb are contained in a face of P whose last coordinate is the constant m
and whose first d − 1 coordinates lie in {0, 1, . . . , k}, by the inductive hypothesis, their
distance is at most k(d − 1). Hence the distance from v to w is at most
k(d − 1) + a + b ≤ k(d − 1) + cv + cw − 2m ≤ k(d − 1) + (M − m) ≤ kd.
If instead cv + cw ≥ m + M , then we may apply an analogous argument using a sequence
v = v0 , v1 , . . . , va of adjacent vertices such that cvi < cvi+1 and cva = M .
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